MAT495_CHAPTER_4

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Part 1 PARTIAL DERIVATIVES MAT 295

51

Chapter 4

Chain Rule &

ImplicitDifferentiationAt the end of this chapter, students should be able to:

Define the chain rule

Perform chain rule

Define implicit differentiation

Perform and compute the implicit differentiation

4.1 Introduction

In the previous chapter, the procedure to determine the second order and

higher order partial derivatives was discussed. In Chapter 4, the chain rule

and implicit differentiation will be extended to function of more than onevariable. These will allow the generation of useful relationships among the

derivatives and partial derivatives of various functions.

4.2 CChhaaiinnRRuullee

The standard chain rule for functions of one variable has been used

throughout in the earlier calculus courses. Its now time to extend the chain

rule out to more complicated situations. Lets review the notation for the

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chain rule for functions of one variable before we move further into functions

of several variables.

If zis a differentiable function ofxandxis a differentiable function of t, then

the chain rule for functions of one variable states that, under assumption, z

becomes a differentiable function of twith

dt

dx

dx

dz

dt

dz

(a) (b)

Figure 4.1

Figure 4.1 illustrates the tree diagram for the chain rule for functions of one

variable. Figure 4.1(a) is a tree diagram in horizontal form where as Figure

4.1(b) is in vertical form.

Theorem

Chain Rule: One Independent Vari able

Let ),,( yxfz where f is a differentiable function of x and y. If )(tgx ,

)(thy , where g and h are differentiable functions of t, then z is a

differentiable function of t,

dt

dy

y

z

dt

dx

x

z

dt

dz

z

x

t

f

g

f(x)

g(t)

t x z

g(t) f(x)

z = f(g(t))

g f

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(a) (b)

Figure 4.2

Theorem

Chain Rule: Two I ndependent Vari ables

Let ),,( yxfz where f is a differentiable function of x and y. If )t,s(gx ,

),( tshy , such that the first partialss

x

,

t

x

,

s

y

, and

t

y

all exist, then

s

z

andt

z

exist and are given by

s

y

y

z

s

x

x

z

s

z

andt

y

y

z

t

x

x

z

t

z

(a) (b)

Figure 4.3

z

y

xx

z

y

z

t

y

s

y

t

s

t

x

t

s

s

x

z

yx

x

z

y

z

t

y

s

y

t s

t

x

t s

s

x

z

t

x

yx

t

x

z

y

z

dt

dy

t

x=g(t)

z=f(x, y)

z = f(g(t), h(t))

g f

y=h(t)

f

x

z

yh

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Proposition

Let ),,( zyxfw be a function defined in region R in space. Suppose that is

a curve in R given parametrically by )(txx , )(tyy , )(tzz , with t = 0

corresponding to ),,( 000 zyx . Then, considering ))(),(),(( tztytxfw as a

function of talong , we have

dt

dz

z

w

dt

dy

y

w

dt

dx

x

w

dt

dw

(1)

That is, the rate of change of w with respect to talong is given by (1) .

Example 1

Finddt

dz, )ln( 22 yxz , tex , tey

State chain rule

dt

dy

y

z

dt

dx

x

z

dt

dz

Figure 4.4

Solut ion

Steps : Using chain rule

Identify the variable Find the derivatives Apply chain rule Substitute in forxand yvariables (or other

variable depending on the problem

z

dt

dx

yx

t

x

z

y

z

dt

dy

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then

yx

w

yzxy

w2

2yz

w

;

and

1dt

dx t

dt

dy2

tdt

dz 1 .

Hence,

tytyzxy

dt

dz

z

w

dt

dy

y

w

dt

dx

x

w

dt

dw

1)2)(2()1( 2

At t= 2,

2 tx 4222 ty 2lnln tz

giving

2ln3220

2

116)4)(2ln)4(22()1(4

dt

dw

Example 3

If )( 22

qpfv , show that 0

p

v

qq

v

p .

Note that v is a function of single variable. Letting 22 qpw then

)(wfv . Figure 4.6 illustrates the vertical tree diagram for )(wfv

Figure 4.6

q

w

dw

dv

q

v

.

p

w

dw

dv

p

v

.

Solut ion

v

dw

dv

qp

wp

w

q

w

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qdw

dv

q

v2.

pdw

dv

p

v2.

dw

dvq2

dw

dvp2

0

22

dw

dvpq

dw

dvpq

p

vq

q

vp

Example 4

The radius of a cylinder increases at the rate of 0.2cm/s while the height

decreases at the rate of 0.5cm/s. Find the rate at which the volume is

changing at the instant when r = 8cm and h= 12cm.

Given that

2.0dt

dr 5.0

dt

dh

while

),(2 hrVhrV

where the radius (r) and height (h) depend on time (t). Hence, the tree

diagram to illustrates the given information is represented by Figure 4.7.

Figure 4.7

dt

dhV

dt

drV

dt

dVhr

dt

dhr

dt

drrh 22

)5.0()8()2.0)(12)(8(2 2 ; substitute the value

s/cm1.20 3

Solut ion

V

dt

dr

hr

t

r

V

dt

dh

h

V

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Example 5

Suppose that 2)cos( yxyw , tex 3 and ty 3sin , finddt

dw

Figure 4.8

dt

dy

y

w

dt

dx

x

w

dt

dw

)).(sin( yxyx

w

yxxyy

w2)).(sin(

tedt

dx 33 tdt

dy3cos3

xyxytxyye

dt

dy

y

w

dt

dx

x

w

dt

dw

t sin23cos3sin3 3

Given that )2ln( 2 yxz ; 4tx , 2ty

(i) Find the partial derivatives fordtdyand

dtdx

yz

xz ,,

.

(ii) Then apply chain rule to finddt

dz.

(iii) Substitute 4tx and 2ty in your answer at (ii).

Solut ion

Warm up exercise

w

dt

dx

yx

t

x

w

dt

dy

y

w

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4.3 IImmpplliicciittddiiffffeerreennttiiaattiioonn

A method called implicit differentiation makes use of the chain rule to

differentiate implicitly defined functions.

Theorem

The Implicit Function Theorem states that if F is defined on an open disk

containing (a,b), where 0),( baF , 0),( baFy , and xF and yF are

continuous on the disk, then the equation 0),( yxF defines yas a function

ofxnear the point (a,b) and the derivative of this function is given by:

y

x

F

F

y

fx

f

dx

dy

Proposition

Suppose that f is a differentiable function of ),,( zyx near the point

),,( 000 zyx and .0),,( 000 zyxfz Then the equation 0),,( zyxf defines z

implicitly as function ofx, y and

z

x

f

f

x

z

and z

y

f

f

y

z

Example 6

Finddx

dyby using implicit differentiation.

a) 1002 233 yxyx

b) xyyx sintan

Steps : Using implicit differentiation

Get a zero on one side of the equal sign andidentify F(x,y)

Find the partial derivatives

A l the im licit functions theorem
http://en.wikipedia.org/wiki/Chain_rulehttp://en.wikipedia.org/wiki/Chain_rule

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a) 1002 233 yxyx

Rewrite

01002),( 233 yxyxyxF

Differentiate with respect tox

xyxFx 43 2

Differentiate with respect to y

22 23 xyFy

Apply theorem

22

2

23

34

xy

xxy

F

F

dx

dy

y

x

b) xyyx sintan ,

0sintan),( xyyxyxf

xyyFx costan

xyxFy sinsec2

xyx

yxy

F

F

dx

dy

y

x

sinsec

tancos2

Example 7

Obtain the equation of the tangent line to the graph 16 33 xxyy at

a point (1, 2).

In general the equation of a tangent line is given by

cmxy

Solut ion

Solut ion

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where mis the slope and it can be obtained by determiningdx

dym

where

y

x

F

F

dx

dy

Since,

16),( 33 xxyyyxF

236 xyFx

xyFy 63 2

xy

xy

F

F

dx

dy

y

x

63

362

2

At (1,2)

)1(6)2(3

)1(3)2(62

2

dx

dy

2

3 This is the slope of the tangent line.

Hence, the line is

2

3

1

2

x

y

)1(2

32 xy

2)1(2

3 xy

)13(2

1 x

Example 8

If 04248984 22 yxyxyx and 0dx

dy, show that 1yx .

4248984),(

22

yxyxyxyxF 888 yxFx

Solut ion

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24188 yxFy

024188

888

yx

yx

F

F

dx

dy

y

x

0888 yx

1yx

Example 9

Given zyxzzzyxf 223 3),,( , find expression forxz and

yz

where

zis defined implicitly as a function of ),( yx by the equation 5),,( zyxf .

Evaluate these at the point (1, 1, 1).

Compute partial derivatives

23zx

f

yzy

f2

22 63 yxzzz

f

Hence,

22

2

63

3

yxzz

z

x

z

and22 63

2

yxzz

yz

y

z

The values at (1, 1, 1) are

10

3

x

z and

5

1

y

z

Solut ion

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Let 16 22 xxyy .

(i) Rewrite the function as 0),( yxF .

(ii) Find the partial derivatives yx FandF .

(iii) Then, obtain the equation of the tangent line to the graph

16 22 xxyy at a point (2,4) by using implicit functions theorem.

Exercise 4

Chain Rule

1. Use the chain rule to find the rate of change of z with respect to t.

a) 323 yxz ; 4tx , 2ty

b) xyxz sincos3 ;t

x 1 , ty 3

c) yxz 22ln ; 4tx , 12 ty

d) 421 xyxz ; tx ln , ty

2. Apply the chain rule to findv

zand

u

z

.

a)v

uyvuxez xy ,2,2

b) vuvuz ,3,tansin

c) ,y

xz vuyuvx sin,cos

d) xyxz tan2 , 22, vuyv

ux

3. Find uz if

a) vusanduvrsr

rsz 2,

22

b) uvwvuyvuxwyxz sin2,,,)2ln( 22

Warm up exercise

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4. If 43yxw and 35,22 tytx , evaluatedt

dw when 1t .

5. Let )2ln( 2 yxZ where rtyandrtx 232 . Findt

Z

when

2,1 rt .

6. If ),( yxfz , where fis differentiable, )(),( thytgx , g(3) = 2, g(3)= 5,

h(3) = 7, h(3)= - 4, 6)7,2( xf , and 8)7,2( yf , finddt

dzwhen t= 3.

Implicit Function

1. Finddx

dyby using implicit partial differentiation:

a) 53 323 yxyx 0cos32 yyx

b) yxyx cossincossin

c) yxy yee 1

d) yxyx 34

2. Finddx

dygiven that 300

24

y

x

x

y.

3. Given that 33 yxz and 122 yx . Determine an expression fordx

dz in

term of x and y.

4. Obtain the equation of the tangent line to the graph 16 22

xxyy at a point(2, 4).

5. If 1 yxexy , evaluatedx

dyat (0,0).

6. Find an expression fordx

dywhen xyyx sintan .

7. Let 10322 yzx , findx

z

and

y

z

.

8. Givenx

yexz 782 , find the rate of change of z with respect to y.

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Miscellaneous Exercises

1. The radius r and height h of a right circular cylinder are increasing at rates of

0.02 cm/min and 0.01 cm/min respectively. Using partial differentiation, find the

rate at which the volume is increasing when r = 4 cm and h = 7 cm.

2. Two straight roads intersect at right angles. Car A, moving on one of the roads,

approaches the intersection at 25km/h and car B, moving on the other road,

approaches the intersection at 30km/h. At what rate is the distance between

the cars changing when A is 0.3 km from the intersection and B is 0.4 km from

the intersection?

3. Suppose that the portion of a tree that is usable for lumber is a right circular

cylinder. If the usable height of a tree increases 2m/year and the usable

diameter increases 3cm/year, how fast is the volume of usable lumber

increasing when the usable height of the tree is 20m and the usable diameter is

30cm?

4. The radius of a right circular cone is increasing at a rate of 1.5 cm/s while the

height is decreasing at a rate of 2 cm/s. If the surface area of the cone is

2 2 2S r r r h , find the rate of change of S when the radius is 4 cm and

the height is 6 cm.

5. As a solid right circular cylinder is heated, its radius r and height h increase;

hence, so does its surface area S. Suppose that at the instant when r = 10 cm

and h = 100 cm, r is increasing at 0.2 cm/min and h is increasing at 0.5 cm/min.

How fast is S increasing at this instant?

6. Use partial differentiation to find the equation of the tangent line to the graph of

0xy3yx 33 at the point (1, 4).

7. Find the slope of a tangent line to the curve 2 210 x y 3 at the point (2, 1).

8. Use formula from partial differentiation to find the slope of the tangent line to the

curve 15 223 xyxyx at any point (x,y).

9. Find the equation of the tangent to the graph of the equation

02)ln( yxxy at the point (2, -1) using partial differentiation.

10. For an implicit function yxxyxyexy 2sin , compute the slope of the

tangent line at point (, 1) by using partial differentiation.

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MAT495_CHAPTER_4